After blast-off, a space shuttle climbs vertically and a radar-tracking dish, located 1000 meters from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blast-off if at that time the velocity of the shuttle is 100 m/sec and the shuttle is 500 meters above the ground?
To find the speed at which the radar dish is revolving 10 seconds after blast-off, you can use the concept of similar triangles.
Let's consider two triangles: Triangle 1, formed by the radar dish, the ground, and the shuttle; and Triangle 2, formed by the radar dish, the shuttle, and its height above the ground.
Triangle 1 is a right triangle, with the hypotenuse being the distance between the radar dish and the launch pad, which is 1000 meters. The height of the radar dish, which we want to find, will be one of the legs of this triangle.
Triangle 2 is also a right triangle, with the hypotenuse being the distance between the radar dish and the shuttle, which we know is 1000 meters. The height of the shuttle, which is 500 meters, will be one of the legs of this triangle.
Since the two triangles are similar, their corresponding sides are proportional. Therefore, we can set up the following equation:
(height of the shuttle) / (distance between the shuttle and the radar dish) = (height of the radar dish) / (distance between the radar dish and the launch pad)
Plugging in the given values:
500 / 1000 = (height of the radar dish) / 1000
Simplifying this equation, we get:
(height of the radar dish) = (500 / 1000) * 1000 = 500
So the height of the radar dish is 500 meters.
Now, let's consider the velocity of the shuttle at this time. We are given that the velocity of the shuttle is 100 m/sec.
Since the radar dish is following the shuttle, it has to revolve in such a way that it maintains a constant line of sight with the shuttle. This means that the time derivative of the angle between the line connecting the radar dish and the shuttle and the horizontal line would give us the angular velocity.
In simpler terms, the angular velocity at which the radar dish is revolving is equal to the derivative of the angle of elevation between the shuttle and the radar dish.
Using trigonometry, we can find this angle:
tan(angle of elevation) = (height of the shuttle) / (distance between the shuttle and the radar dish)
tan(angle of elevation) = 500 / 1000 = 0.5
Taking the inverse tangent of both sides, we get:
angle of elevation = arctan(0.5) ≈ 26.6°
Now, we can find the angular velocity (revolutions per second) using the formula:
angular velocity = (tangent of the angle of elevation) * (velocity of the shuttle) / (distance between the shuttle and the radar dish)
angular velocity = tan(26.6°) * 100 / 1000 = 0.5 * 0.1 = 0.05 revolutions per second
So, the radar dish is revolving at a speed of 0.05 revolutions per second 10 seconds after blast-off.
To find the speed at which the radar dish is revolving, we can use the concept of relative velocity.
The velocity of the shuttle and the velocity of the dish are related as follows:
Velocity of the dish = Velocity of the shuttle + Velocity of rotation
In this case, the velocity of the shuttle is given as 100 m/sec, and we need to find the velocity of rotation.
First, let's find the distance between the dish and the shuttle at the given time (10 sec after blast-off).
The shuttle has traveled a vertical distance of 500 meters in these 10 seconds, so it is at a height of 500 meters. The distance between the dish and the shuttle is constant at 1000 meters.
Now, let's find the distance between the dish and the shuttle horizontally. Since the shuttle is climbing vertically, the horizontal distance remains the same.
Using the Pythagorean theorem, we can find the horizontal distance:
Horizontal distance = sqrt((Total distance)^2 - (Vertical distance)^2)
= sqrt((1000m)^2 - (500m)^2)
= sqrt(1,000,000m^2 - 250,000m^2)
= sqrt(750,000m^2)
≈ 866m
So, at the given time, the horizontal distance between the dish and the shuttle is approximately 866 meters.
Now, we can calculate the velocity of rotation using the formula:
Velocity of rotation = (Change in horizontal distance) / Time
= (Change in horizontal distance) / (Time interval)
Since the horizontal distance remains constant and the time interval is 10 seconds:
Velocity of rotation = 0 / 10
= 0 m/s
Therefore, the speed at which the radar dish is revolving 10 seconds after blast-off is 0 m/sec.
h = height = 500 + 100 t
b = base = 1000
A = elevation angle. we want dA/dt
tan A = h/b = 100(5+t) / 1000 = .5 +.1 t
(1 /sec^2 A) dA/dt = 0 + .1
dA/dt = .1 sec^2 A
at this time ratio of b to h is 2/1
so hypotenuse is sqrt 5
cos A = 2/sqrt 5
sec A = sqrt 5/2
sec^2 A = 5/4
so
dA/dt = .1 (5.4) = .54 radians/second
or
31 degrees per second
sec A = 1/cos A = 1/(1000/(1000^2